Unidad IV Eqsolubilidadalumnos

8/14/2019 Unidad IV Eqsolubilidadalumnos

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UNIDAD IV. EQUILIBRIOS DE

SOLUBIBLIDAD


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Reacciones de precipitacin

Los iones solubles sepueden combinar para

formar un compuesto

insoluble.

Se produce laprecipitacin.

Ag+(aq) + Cl-(aq) AgCl(s)


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Ag+(aq) + NO3-(aq) + Na+(aq) + I-(aq)

AgI(s) + Na+(aq) + NO3-(aq)

Iones espectadores

Ecuacin inica neta

AgNO3(aq) +NaI (aq) AgI(s) + NaNO3(aq)

Reaccin de precipitacin global:

Ecuacin inica neta:

Ag+(aq) + I-(aq) AgI(s)

Ecuacin inica neta:


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Reglas de solubilidad

Compuestos solubles:

Li+, Na+, K+, Rb+, Cs+ NH4+

NO3- ClO4

- CH3CO2-

Todos los de losmetales alcalinos y elion

amonio.

Nitratos, percloratosy acetatos


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Reglas de solubilidad

Hidrxidos y sulfuros HO-, S2- Excepto los metales alcalinos y los iones amonio.

Los sulfuros de los alcalinotrreos son solubles. Los hidrxidos de Sr2+ y Ca2+ son un poco solubles.

Carbonatos y fosfatos CO32-, PO4

3- Excepto los de los metales alcalinos y las sales de amonio.

Compuestos insolubles:


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Temperature and Solubility

Solid solubility and temperature

solubility increases withincreasing temperature

solubility decreases with

increasing temperature

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Fractional crystallizationis the separation of a mixture of substances into pure components

on the basis of their differing solubilities.

Suppose you have 90 g KNO3

contaminated with 10 g NaCl.

Fractional crystallization:

1. Dissolve sample in 100 mL ofwater at 600C

2. Cool solution to 00C

3. All NaCl will stay in solution (s =

34.2g/100g)

4. 78 g of PURE KNO3will

precipitate (s = 12 g/100g). 90

g 12 g = 78 g

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Temperature and Solubility

Gas solubility and temperature

solubility usually decreaseswith increasing

temperature

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Solubility Equilibria

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AgCl (s) Ag+(aq)+ Cl-(aq)

Ksp= [Ag+][Cl-] Kspis the solubility product constant

MgF2(s) Mg2+(aq)+ 2F-(aq) Ksp= [Mg

2+][F-]2

Ag2CO3(s) 2Ag+(aq)+ CO3

2-(aq) Ksp= [Ag+]2[CO3

2-]

Ca3(PO4)2(s) 3Ca2+(aq)+ 2PO4

3-(aq) Ksp= [Ca2+]3[PO3

3-]2

Dissolution of an ionic solid in aqueous solution:

Q= Ksp Saturated solution

Q< Ksp Unsaturated solution No precipitate

Q> Ksp Supersaturated solution Precipitate will form


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Molar solubility(mol/L) is the number of moles of solute dissolved in 1 L of a saturated

solution.

Solubility(g/L)is the number of grams of solute dissolved in 1 L of a saturated solution.

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What is the solubility of silver chloride in g/L ?

AgCl (s) Ag+(aq)+ Cl-(aq)

Ksp= [Ag+][Cl

-]Initial (M)

Change (M)

Equilibrium (M)

0.00

+s

0.00

+s

s s

Ksp= s2

s= Ksp

s= 1.3 x 10-5[Ag+] = 1.3 x 10-5M [Cl-] = 1.3 x 10-5M

Solubility of AgCl =1.3 x 10-5 mol AgCl

1 L soln

143.35g AgCl

1 mol AgCl

x = 1.9 x 10-3g/L

Ksp= 1.6 x 10-10

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If 2.00 mL of 0.200 MNaOH are added to 1.00 L of 0.100 MCaCl2, will a

precipitate form?

16.6

The ions present in solution are Na+, OH-, Ca2+, Cl

-.

Only possible precipitate is Ca(OH)2(solubility rules).

Is Q> Kspfor Ca(OH)2?

[Ca2+]0= 0.100 M [OH-]0= 4.0 x 10-4M

Ksp= [Ca2+][OH

-]2= 8.0 x 10-6

Q= [Ca2+]0[OH-]0

2 = 0.10 x (4.0 x 10-4)2= 1.6 x 10-8

Q< Ksp No precipitate will form


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The Common Ion Effect and Solubility

The presence of a common ion decreases the solubility of the

salt.

What is the molar solubility of AgBr in (a) pure water and (b) 0.0010 MNaBr?

AgBr (s) Ag+(aq)+ Br-(aq)

Ksp= 7.7 x 10-13

s2= Ksp

s= 8.8 x 10-7

NaBr (s) Na+ (aq)+ Br-(aq)

[Br-] = 0.0010 M

AgBr (s) Ag+(aq)+ Br-(aq)

[Ag+] = s

[Br-] = 0.0010 + s0.0010

Ksp= 0.0010 x s

s= 7.7 x 10-10

16.8


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Complex Ion Equilibria and Solubility

A complex ionis an ion containing a central metal cation bonded to one or more molecules

or ions.

Co2+(aq)+ 4Cl-(aq) CoCl4(aq)2-

Kf=[CoCl4]

[Co2+][Cl-]4

2-

Theformation constant or stability constant (Kf)is the equilibrium constant for the

complex ion formation.

Co(H2O)6 2+ CoCl42-

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Kfstability ofcomplex


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16.10


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16.11


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Qualitative

Analysis of

Cations

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Flame Test for Cations

lithium sodium potassium copper

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